100 POINTS! PLEASE HELP!
2 answers:
You might be interested in
Substitute 
, so that
![\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dx%5E2%7D%3D%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5B%5Cdfrac1x%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dz%7D%5Cright%5D%3D-%5Cdfrac1%7Bx%5E2%7D%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dz%7D%2B%5Cdfrac1x%5Cleft%28%5Cdfrac1x%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dz%5E2%7D%5Cright%29%3D%5Cdfrac1%7Bx%5E2%7D%5Cleft%28%5Cdfrac%7B%5Cmathrm%20d%5E2y%7D%7B%5Cmathrm%20dz%5E2%7D-%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dz%7D%5Cright%29)
Then the ODE becomes
which has the characteristic equation
with roots at 
. This means the characteristic solution for 
is
and in terms of
, this is
From the given initial conditions, we find
so the particular solution to the IVP is
Then the ODE becomes
which has the characteristic equation
and in terms of
From the given initial conditions, we find
so the particular solution to the IVP is
Perfect Square Trinomials have first and last terms that are positive perfect squares. And the middle term is
twice the product of the square roots of the first and third terms.
First term:
n^2
Square root = n
Third term:
9
Square root = 3
9 * 1 = 9
9 * 2 = 18
18
-9
Therefore it is not a perfect square trinomial.
First term:
n^2
Square root = n
Third term:
9
Square root = 3
9 * 1 = 9
9 * 2 = 18
18
Therefore it is not a perfect square trinomial.